Question: Let $a^2=\frac{16}{44}$ and $b^2=\frac{(2+\sqrt{5})^2}{11}$, where $a$ is a negative real number and $b$ is a positive real number. If $(a+b)^3$ can be expressed in the simplified form $\frac{x\sqrt{y}}{z}$ where $x$, $y$, and $z$ are positive integers, what is the value of the sum $x+y+z$?
Solution: First we solve for $a$ and $b$. $$a=-\sqrt{\frac{16}{44}}=-\frac{\sqrt{16}}{\sqrt{44}}=-\frac{4}{2\sqrt{11}}=-\frac2{\sqrt{11}}$$$$b=\sqrt{\frac{(2+\sqrt{5})^2}{11}}=\frac{2+\sqrt{5}}{\sqrt{11}}$$Now we solve for $(a+b)^3$. \begin{align*}(a+b)^3&=\left(-\frac2{\sqrt{11}}+\frac{2+\sqrt{5}}{\sqrt{11}}\right)^3=\left(\frac{\sqrt{5}}{\sqrt{11}}\right)^3=\frac{\sqrt{5^3}}{\sqrt{11^3}}\\
&=\frac{5\sqrt{5}}{11\sqrt{11}}=\frac{5\sqrt{5}}{11\sqrt{11}}\cdot\frac{\sqrt{11}}{\sqrt{11}}=\frac{5\sqrt{55}}{121}
\end{align*}So, $x+y+z=5+55+121=\boxed{181}$.